Physics 2101 Physics 2101 Section 3Section 3Section 3Section 3

May 7May 7thth: Chap. 20: Chap. 20Announcements:Announcements:•• Final Exam: May 11Final Exam: May 11th th

(Tuesday), 7:30 AM at Howe(Tuesday), 7:30 AM at Howe--Russell 130Russell 130

•• Make up Final: May 15Make up Final: May 15th th

(Saturday) 7:30 AM at Nicholson (Saturday) 7:30 AM at Nicholson 119119

•• Final Exam for those who need Final Exam for those who need extended time will be at extended time will be at Nicholson 109 Nicholson 109 R i d 2PM R i d 2PM

Chapter 20Chapter 20——Irreversible processesIrreversible processes

•• Review: Saturday 2PM at Review: Saturday 2PM at Nicholson 130Nicholson 130

Class Website:Class Website: http://www.phys.lsu.edu/classes/spring2010/phys2101http://www.phys.lsu.edu/classes/spring2010/phys2101--3/3/http://www.phys.lsu.edu/~jzhang/teaching.htmlhttp://www.phys.lsu.edu/~jzhang/teaching.html

History of grades in 2101 for last 5 years:

I will post on Webassign you standing in the class ,i.e. 15 out of 59. This will give

The rough guide lines are The rough guide lines are %%

the class ,i.e. 15 out of 59. This will give you a way to determine where you are

A: < 18 %A: < 18 %B: 17 B: 17 –– 25 %  25 %  (with A+B < 40%)   (with A+B < 40%)   C: 33C: 33 42 %42 %C: 33C: 33‐‐42 % 42 % 

Entropy: Reversible processes

1) For reversible process:  

ΔScycle,rev = 0 =dQT

∫

2) For isothermal gas  process:  

ΔSisothermal =QT

ΔE int = 0 ⇒ Q = W

lVf

⎛ ⎜

⎞ ⎟ ΔS R l

Vf⎛ ⎜

⎞ ⎟

3) For liquids/solids ‐ heat transfer:(assuming no volume change)

T W = nRT ln f

Vi⎝ ⎜

⎠ ⎟

isothermal

⇒ ΔSisothermal = nR ln f

Vi⎝ ⎜

⎠ ⎟

ΔSliquid / solidheat transfer

= S f − Si = mc lnTf

Ti

⎛

⎝ ⎜

⎞

⎠ ⎟

Qheat capacity = mc Tf −Ti( )dQheat capacity = mc dT( )

4) For phase transition:   

Q = mL ΔS = S − S =mLQphase transition = mL ΔSphase transition = S f Si =T

Heat Engines and Entropy‐Easy to produce thermal energy by doing work.  How?

‐Much harder to get work from thermal energy  ⇒ engine

Conservation of energy|QH| = |QL| + |W|H L|QH| = heat added|QL| = heat released

What is the thermal efficiency, ε, of an engine?

energy we get Wε =

energy we getenergy we pay for

=QH

Q − Q Q Perfect engine, ε=1ε =

QH QL

QH

= 1−QL

QH

How close to ideal can we get?  Carnot Engineε =

QH − QL = 1QL

1) a → b: Isothermal expansion

ΔE int = 0 ⇒ QH = W = nRTH ln Vb

V⎛

⎝⎜

⎞

⎠⎟ > 0

All processes are reversible and no wasteful energy transfers occur ε =QH

= 1−QH

2) b → c: adiabatic expansion

int QH H Va⎝ ⎜

⎠ ⎟

ΔSH = QH

TH

= nR ln Vb

Va

⎛

⎝ ⎜

⎞

⎠ ⎟ > 0 positive

3) c → d: Isothermal compressionV⎛ ⎞

Q = 0 ⇒ ΔSb→c = 0

0 = ΔScycle = ΔSH + ΔSL

ΔE int = 0 ⇒ QL = W = nRTL ln Vd

Vc

⎛

⎝ ⎜

⎞

⎠ ⎟ < 0

ΔSL = QL

TL

= nR ln Vd

Vc

⎛

⎝ ⎜

⎞

⎠ ⎟ < 0 negative

0 =QH

TH

−QL

TL

4) d → a: adiabatic compression

Q = 0 ⇒ ΔSd →a = 0

εcarnot = 1−TL

TH

All Carnot engines operating between two constant temperatures TH & TL have the same efficiency.

An irreversible engine is less efficient  

Problem #1 ε =WQH

=QH − QL

QH

= 1−QL

QH

εcarnot =1−TL

TH

Imagine a Carnot engine  that operates between the temperatures of TH = 850 K and TL = 300 K.  The engine performs 1200 J of work each cycle, which takes 0.25 s.

a) What is the efficiency of this engine?

b) What is the average power of this engine?

The efficiency, ε, of a Carnot engine is ONLY determined by the ratio TL/TH:

Power is found from Work done per cycle/time per cycle:

ε = 1−TL

TH

= 1−300 K850 K

≅ 65%

P = W = 1200 J( )= 4800 W

c) How much energy |QH| is extracted as heat from the high temperature reservoir every cycle?

Power is found from Work done per cycle/time per cycle:

For any engine, the efficiency, ε, is defined as the work the engine does per cycle 

Pave = Δt = 0.25 s( )= 4800 W

d) How much energy |QL| is delivered as heat to the low temperature reservoir every cycle?

y g y g p ydivided by the energy it absorbs as heat per cycle:

ε =WQH

→ QH =Wε

=1200 J65%

≅ 1855 J

F C t i f th 1st L f th th t th t h t t f

e) What is the entropy change ΔS of the working substance for the energy transfer to it from 

For a Carnot engine, from the 1st Law of thermo says that the net heat transfer per cycle is equal to the  net work done: W = QH − QL → QL = QH − W = 1855 J −1200 J = 655 J

the high‐temperature and low‐temperature reservoir?Along the hot and cool isotherms, the entropy changes are: 

ΔSL =QL

TL

=−655 J300 K

= −2.18 J/K ΔSH =QH

TH

=1855 J850 K

= 2.18 J/K → ΔStot = 0

Otto cycle: 4‐stroke internal combustion engine

a→ b : adiabatic compression stroke b → c : isochoric expansion due to fuel ignition

c → d : adiabatic expansion stroke d → a : isochoric compression due to exhaust valve

What is the efficiency of the Otto cycle?

Assume the compression ratio of 4:1 (V4=4V1), the gas mixture can be regarded as 

id l t ian ideal gas monatomic gas:Theoretical : 40%Reality : 20%

Problem

Problem 20‐29 One mole of a monatomic ideal gas is taken through the

ε =WQH

=QH − QL

QH

= 1−QL

QH

Problem 20 29 One mole of a monatomic ideal gas is taken through the reversible cycle shown. Vc=8.00Vb , pb=10.0 atm and Vb=10‐3m3.) What’s the energy added to the gas as heat? b) the energy leaving the gas as heat, c) What’s the net work done by the gas?  d) What’s the efficiency of the cycle?the cycle?

-1 -1 1 1b b c c 1

At the state b: so that /

5 1: T V T V ( ), so that ( ) ( )3 8 8

b b b b b b

p b bc b b

PV RT T PV RC V Tb c T T TC V

γ γ γ γγγ − −

−

= =

⇒ = = = = = =

Energy added to the gas as heat:

3 8 8

( ) : ; so ( ) ( )8 8

v c

a a a b c ba c a c c

c c c c

C VV T V V T Tc a p p T T TV T V V γ⇒ = = = = = =

Energy leaving from the gas as heat:Energy added to the gas as heat:3 ( )2ab V ab b aRQ nC T T T= Δ = −

gy g g5 ( )2ca p ca c aRQ nC T T T= Δ = −

Net work:( 2 )

( ) ( )net bc ca b a cW W W R T T T

W E R T T= + = + −

Δ( ) ( )( )

bc in bc b c

ca a ca ac a c

W E R T TW p V R T R T T

= − Δ = −= Δ = Δ = −

Problem #1Entropy change with phase transition

What is the entropy change of a 12.0 g ice cube that melts completely in a bucket of water whose temperature is just above the freezing point of water?

Entropy change with phase transition

ΔS = S f − Si =dQTi

f

∫How do we find the change in entropy?

Qphase transition = mLIn this case, what is the heat transferred to the system??

Plugging this in, we can calculate the entropy change:  (this is a reversible isothermal process!!)

ΔSphase change =mLT

Thus: 

ΔSice =12 g( ) 333 J/g( )

273 K( )= 14.6 J/K ≥ 0

( )

Problem #2

A 50 0 g block of copper whose temperature is 400 K is placed in an insulating box with a 100 g

Solid: entropy change with temperature

A 50.0 g block of copper whose temperature is 400 K is placed in an insulating box with a 100 gblock of lead whose temperature is 200 K (cCu=386 J/kg/K & cPb=128 J/kg/K )a) What is the equilibrium temperature of the two‐block system?

( )∑ ( )We know from Chapt. 19 that in an insulating box, the net heat flow is zero, thus:

Review question

b) What is the change in internal energy of the two block system between the initial state and the

Q = 0 = mCucCu Tf −Ti,Cu( )+∑ mPbcPb Tf −Ti,Pb( )Tf =

mCucCuTi,Cu + mPbcPbTi,Pb

mCucCu + mPbcPb

= 320 K

b)    What is the change in internal energy of the two‐block system between the initial state and theequilibrium state?   

Since the system (Cu + Pb) is thermally insulated and no work is done, the change in the internal energy is zero!

c)  What is the change in the entropy of the two‐block system?

ΔSrev = S f − Si =dQTi

f

∫Knowing                                               what is the heat transfer?

dQheat capacity = mc dT( )

ΔS =mc dT( )f

∫ = mc lnTf

⎛ ⎜

⎞ ⎟

ΔStot = ΔSCu + ΔSPb = mCucCu ln 320 K400 K

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + mPbcPb ln 320 K

200 K⎛ ⎝ ⎜

⎞ ⎠ ⎟

ΔSTi

∫ mc lnTi⎝

⎜ ⎠ ⎟

= − 4.3 J/K + 6.0 J/K = 1.7 J/K (positive!!)

Problem #3Gas: entropy change during process

An ideal gas undergoes an isothermal expansion at 77°C from 1.3 l to 3.4 l.  Then entropy change during this process is 22 J/K.  How many moles of gas are there?

What is the change in entropy during an isothermal process?

ΔS dQf

∫ Q ΔE 0ΔSisothermal =QTi

∫ =QT

ΔE int,isothermal = 0

What do we know from 1st Law of Thermo?

ΔE int,isothermal = 0 ⇒ Q = W = pdV∫ ⇒ Q = nRT lnVf

Vi

⎛

⎝ ⎜

⎞

⎠ ⎟

Putting together, we find that:

n =ΔSisothermal

Vf⎛ ⎞

=22 J/K

8 31 / l( )l 3.4 l⎛ ⎜

⎞ ⎟

= 2.75 molesR ln

Vf

Vi

⎛

⎝ ⎜

⎞

⎠ ⎟ 8.31 J/mole ⋅ K( )ln 3.4 l

1.3 l⎛⎝ ⎜

⎞⎠ ⎟

Problem #4Gas: entropy change during process

One mole of an ideal monatomic gas is taken through the cycle shown.  

a) How much work is done by the gas in going from state a to state c along abc?

Work done is: Wnet = Wa →b + Wb →c = p0dVV0

4V0

∫⎡

⎣ ⎢

⎤

⎦ ⎥

isochoric

+ pdV4V0

4V0

∫⎡

⎣ ⎢

⎤

⎦ ⎥

isobaric

Wnet = Wa →b + Wb →c = 4V0 − V0( )p0 + 0 = 3V0 p0

What are the changes in internal energy and entropy going:b) from b to c ?

Internal energy is given by change in temperature: The entropy is found from: 

dQf

∫dE + W[ ]f

∫ΔSb →c =dQTi

f

∫ =dE int + Wby[ ]

Ti

f

∫

=32 nRdT + 0[ ]

Ti

f

∫⎡

⎣ ⎢

⎤

⎦ ⎥ = 3

2 nR lnTf

Ti

⎛

⎝ ⎜

⎞

⎠ ⎟

isochoric

ΔE int,b →c = 32 nRΔT = 3

2 nR Tf −Ti( )

= 32 nR 2p0 ⋅ 4V0

nR−

p0 ⋅ 4V0

nR⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 6p0V0

c) through one complete cycle ?

Ti⎣ ⎦ isochoricTi⎝ ⎠ isochoric

exp ansion

= 32 R ln

Tf

Ti

⎛

⎝ ⎜

⎞

⎠ ⎟ = 3

2 R ln 8p0V0

4 p0V0

⎛

⎝ ⎜

⎞

⎠ ⎟

= 32 R ln 2( )

For a complete cycle:ΔE int,cycle = ΔSrev ,cycle = 0

19-17: Container A holds an ideal gas at a pressure of 5.0 x 105Pa at a T= 300 K. It is connected to a thin tube and a closed valve to container b with four times theand a closed valve to container b, with four times the volume of A. Container B holds the same ideal gas at a pressure of 1.0 x 105 Pa and T=400K. The valve is opened to allow the pressures to equalize, but the temperature of each container is maintained. What then is the pressure in the two containers?

Use ideal gas Law n = na + nB =VA

RpA

TA

+4 pB

TB

⎛

⎝⎜⎞

⎠⎟= Const.

After valve is openedp 'A =

Rn 'A TA

VA

= p 'B

Rn 'B TBor n 'B =

4TA

T⎛

⎝⎜⎞

⎠⎟n 'A

n = n ' + n ' = n ' 1+4TA⎛

⎜⎞⎟ = n + n =

VA pA +4 pB⎛

⎜⎞⎟

p 'B =Rn B TB

4VA

TB⎝⎜ ⎠⎟

n ' =V pA / TA + 4 pB / TB( )

n = n A + n B = n A 1+TB⎝⎜ ⎠⎟

= na + nB =R TA

+TB⎝⎜ ⎠⎟

n A =R 1 + 4TA / TB

Final AnswerFinal Answer

' RTpA + pB

4TA

Tp ' =n 'A RTA

VA

=A B TB

1 +4TA1 +TB